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Solar Cells.

  1. Nov 14, 2011 #1
    My son is a 3rd grader and wanted to do a science fair project on solar cells.
    So we devised an experiment to see if reflecting light with a mirrow would increase power
    output. We constructed a circuit using an amp-meter, solar cell, and 10Kohm resistor.
    He hypothesized that reflecting sun light (in addition to the incident sun light) would
    create more power; I agreed with his hypothesis. We were surprised that the measured
    power was less:

    Direct sun light only: 360 micro-amps
    Plus reflected sun light: 340 micro-amps

    Can anybody (in terms a 3rd grader would understand) explain why this is?
    BTW, the results are very reproducable.
  2. jcsd
  3. Nov 14, 2011 #2
    Welcome to physics forums!

    Very good hypothesis on your son's part. I have a feeling that you're blocking more light than you're reflecting. Remember, even if you block the sun with your hand stretched out, there is still light coming to your eyes from all over the sky due to scattering. This is by no means an insignificant source of light (after all, you can read under a tree, can't you?).

    Feel free to post a picture or something so that we can confirm.
  4. Nov 14, 2011 #3
    Thank you CMOS for your response.

    We are being very careful not to block any of the incident light. Being that it is November
    and we are located in Texas, the angle of the sun is such that we can easily place the mirrow
    on the table and tilt it such that light reflects on the solar cell without the mirror being over the cell
    at all.

    Thanks again,

  5. Nov 14, 2011 #4
    I'm surprised at the result yet when I checked Wikipedia I did not see any commerical solar cells using reflected light...but that may be purely economics.

    A typical article is here: http://en.wikipedia.org/wiki/Solar_power

    It seems concentrating sunlight IS used for direct heating to produce power...that is thermal energy. So we know reflecting devices definitely concentrate sunlight in the infrared specturm. I don't know what frequency range, if any, is most important for photovoltaic power in solar cells.

    I assume you "turned the mirror on and off" several times and achieved the same reduced current result. Otherwise variations in sunlight could be the culprit...say clouds for example.
  6. Nov 14, 2011 #5
    To clarify, the experiment was only performed on clear sunny days (no clouds affecting the sun light).
  7. Nov 14, 2011 #6
    Hi there,

    there is only one explanation that comes to mind as a possibility. You are most likely heating up the solar cell, and when they get hot, they lose efficiency. A good way to test this would be to set up a little fan to cool the cell a bit. Also, attaching the cell (if possible) to a piece of aluminum or any heat sink will work.

    The physics behind it is fairly in depth, but it has to do with something called the band gap, which represents the voltage that electrons gain when crossing the solar cell junction. As the solar cell heats up, this band gap gets smaller meaning that each electron making the jump will gain less energy. There is an upside to this as well because a smaller band gap means a smaller jump, and that the photons from the sunlight will more easily supply the energy to make the leap.

    EDIT: I feel bad for that explanation because your son is only in grade 3, and my explanation barely makes sense to me :smile:. So I'm trying to figure out a really good analogy for you guys. If I come up with one I'll post it
    Last edited: Nov 14, 2011
  8. Nov 14, 2011 #7
    Measuring the power of a solar cell with a voltmeter (and an am-meter in series with a resistance is just a low-quality voltmeter) tells you little about the power produced.

    The voltage it can deliver depends little on the amount of sunlight, and it will go down a little with increasing temperature.

    The current produced will increase if you increase the amount of sunlight, so
    just hooking up the ammeter (at a higher setting) to the cell will give a much
    clearer result.

    To make an accurate measurement of the power the cell can deliver, you'll need a varying load resistance, and measure both the voltage and the current delivered and found out at what value of the load resistor must be to get the maximum amount of power in the load.

    There's plenty of stuff about concentrating light on a solar cell, such as here:
    http://en.wikipedia.org/wiki/Concentrated_photovoltaics" [Broken]
    Last edited by a moderator: May 5, 2017
  9. Nov 14, 2011 #8
    Hi again,

    I was doing some thinking, are you just measuring the current across the resistor? I'm not sure that this will suffice. This isn't adding up to me because as you increase the sunlight, you increase the temperature, and you increase the current. It is the voltage that decreases as a cause of temperature.

    Power = Voltage * Current, so you will need to multiply the current you see by the voltage across the resistor to actually find power.

    This website shows how to better characterize a solar cell. Admittedly most of this is inappropriate for a third grader, so its up to you to figure out what you want to do.
  10. Nov 14, 2011 #9
    Thanks to everybody that has responded to this post; I think we may be on to something
    with the drop in effiency when temperature goes up theory.

    The reason we're only measuring current is that power delivered to the load (resistor)
    is P = I * I * R. So an increase in current indicates more power being created. The
    experiment is a comparison of methods with the method type being the only variable
    (At his level, only 1 variable is allowed). A 10Kohm resistor was chosen arbitrarily;
    it's a common value and I had some already.


  11. Nov 14, 2011 #10


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    Staff: Mentor

    What is the configuration of your mirror? Are you bouncing the direct sunlight from one mirror to another and then to the solar cell? That should get the most light on the solar cell. I'm guessing that simply using one mirror won't do much because you won't be reflecting direct sunlight. (You should have the solar cell directly facing the sun as well)
  12. Nov 15, 2011 #11
    There is only one mirrow that is abutting the solar cell and angled to place as much of
    the reflected light as possible on the cell. The cell is placed flat on the table; it is not
    angled to be perpendicular to the sun light. However, no noticable difference in
    current is observed when the cell is angled towards the sun (measured without the mirror).
    I believe the solar cell is operating in a saturated mode; thus no difference in the fore-mentioned

  13. Nov 15, 2011 #12
    Saturation can definitely occur, you should also be aware that larger percentages of non-direct light will be reflected by the cell as opposed to direct light. Have you tried a magnifying glass yet? It would seem that it would produce more measureable and consistent results.
  14. Nov 15, 2011 #13
    I have wonder if what you're seeing is related to the internal resistances or other non-idealities of the solar cell itself.

    Try the following:

    1) Take out the resistor and connect the ammeter directly to the solar cell. The current measured should definitely increase with more light.

    2) If you have a voltmeter, try the same as the above; the voltage measured should also definitely increase with more light. Assuming this works, I would even suggest using this as the result of your son's project. Most people know of 1.5-V (AA and AAA) and 9-V batteries; therefore increasing the light makes for a "bigger" battery.

    3) Try your original experiment, but with different resistors. If indeed your original observations were due to internal non-idealities, then it is possible that other resistors will give the results that you originally expected.
  15. Nov 16, 2011 #14


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    Staff: Mentor

    Try it early in the day when the heating will be less pronounced.
  16. Nov 16, 2011 #15
    Yes, the solar cell should show more power when the reflected light from the mirror is added and not less. This is a well established fact and a crucial factor in the rise of concentrated solar photovoltaic power. I, however have no explanation for the results that you claim.
  17. Nov 16, 2011 #16
    It seems likely that this is the problem, obviously some of the sunlight is being cut off, resulting in the lowering of output. Try to position the mirror so that it reflects light but is as far from the cell as possible in order to ensure that it donsn't impede any direct sunlight.
  18. Nov 16, 2011 #17


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    Staff: Mentor

    Try using two mirrors instead of one and pointing the cell at the sun and see if anything changes. Bounce the light off one mirror that is nearly facing the sun head on and off the 2nd mirror onto the cell.
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